Cant understand what are they doing in this part of a solution

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The discussion revolves around proving inequalities involving the limits superior (limsup) and limits inferior (liminf) of sequences x_n and y_n. The initial proof establishes that limsup x_n + limsup y_n is greater than or equal to limsup(x_n + y_n). The next step is to demonstrate that limsup(x_n + y_n) is also greater than or equal to liminf x_n + limsup y_n. A key point made is that limsup(-x_n) equals -liminf(x_n), which leads to the inequality limsup(x_n + y_n) + limsup(-x_n) being greater than limsup y_n. The example provided illustrates that these values can differ, reinforcing that the inequalities do not always yield equalities.
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x_n and y_n are bounded
the first part was proved
we have that
limsup x_n+limsup y_n>=limsup(x_n+y_n)
now we need to prove that:
limsup(x_n+y_n)>=liminf x_n +limsup y_n

as n->infinity
the say:
limsup(-x_n)=-liminf(x_n)

then they say
limsup(x_n+y_n)-liminf x_n=limsup(x_n+y_n)+limsup (-x_n)>=limsup y_n

so by putting liminf x_n on the other side we get
imsup(x_n+y_n)>=liminf x_n +limsup y_n

why
limsup(x_n+y_n)+limsup (-x_n)>=limsup y_n

it should be equal sign
why its bigger ??
 
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Because in general, they're not equal. For instance, let x_n = 0 if n even, 1 if n odd. Then lim sup (-x_n) = 0, and lim sup(x_n) = 1. Now let y_n = 0 for all n. Then we have 1 = lim sup (x_n+y_n) + lim sup(-x_n) > lim sup y_n = 0.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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